47395606 fundamental of microelectronics bahzad razavi chapter 4 solution manual (1)

7/28/2019 47395606 Fundamental of MIcroelectronics Bahzad Razavi Chapter 4 Solution Manual (1)

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4.4 According to Equation (4.8), we have

IC =AEqDnn

2

i

NBWB

eVBE/VT 1

1

WB

We can see that ifWB increases by a factor of two, then IC decreases by a factor of two .


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4.11

VBE = 1.5 V IE(1 k)

1.5 V IC(1 k) (assuming 1)

= VT ln

IC

IS

IC = 775 A

VX IC(1 k)

= 775 mV


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4.12 Since we have only integer multiples of a unit transistor, we need to find the largest number thatdivides both I1 and I2 evenly (i.e., we need to find the largest x such that I1/x and I2/x are integers).This will ensure that we use the fewest transistors possible. In this case, its easy to see that we shouldpick x = 0.5 mA, meaning each transistor should have 0.5 mA flowing through it. Therefore, I1 shouldbe made up of 1 mA/0.5 mA = 2 parallel transistors, and I2 should be made up of 1.5 mA/0.5 mA = 3parallel transistors. This is shown in the following circuit diagram.

VB

+

I1 I2

Now we have to pick VB so that IC = 0.5 mA for each transistor.

VB = VT ln

ICIS

= (26 mV) ln

5 104 A

3 1016 A

= 732 mV


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4.15

VB VBER1

= IB

=IC

IC = R1

[VBVT ln(IC/IS)]

IC = 786 A


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4.17 First, note that VBE1 = VBE2 = VBE.

VB = (IB1 + IB2)R1 + VBE

=R1

(IX + IY) + VT ln(IX/IS1)

IS2

=

5

3IS1

IY =5

3IX

VB =8R13

IX + VT ln(IX/IS1)

IX = 509 A

IY = 848 A


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4.21 (a)

VBE = 0.8 V

IC = ISeVBE/VT

= 18.5 mA

VCE = VCC ICRC

= 1.58 V

Q1 is operating in forward active. Its small-signal parameters are

gm = IC/VT = 710 mS

r = /gm = 141

ro =

The small-signal model is shown below.

B

r

+

v

E

gmv

C

(b)

IB = 10 A

IC = IB = 1 mA

VBE = VT ln(IC/IS) = 724 mV

VCE = VCC ICRC

= 1.5 V


gm = IC/VT = 38.5 mS

r = /gm = 2.6 k

ro =


B

r

+

v

E

gmv

C


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(c)

IE =VCC VBE

RC=

1 +

IC

IC =

1 +

VCC VT ln(IC/IS)

RC

IC = 1.74 mA


VCE = VBE = 739 mV



r = /gm = 2.6 k

ro =


B

r

+

v

E

gmv

C


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4.22 (a)

IB = 10 A

IC = IB = 1 mA


VCE = VCC IE(1 k)

= VCC 1 +

(1 k)

= 0.99 V



r = /gm = 2.6 k

ro =


B

r

+

v

E

gmv

C

(b)

IE =VCC VBE

1 k=

1 +

IC

IC =

1 +

VCC VT ln(IC/IS)

1 k

IC = 1.26 mA


VCE = VBE = 730 mV



r = /gm = 2.07 k

ro =



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B

r

+

v

E

gmv

C

(c)

IE = 1 mA

IC =

1 + IE = 0.99 mA


VCE = VBE = 724 mV



r = /gm = 2.63 k

ro =


B

r

+

v

E

gmv

C

(d)

IE = 1 mA

IC =

1 + IE = 0.99 mA


VCE = VBE = 724 mV



r = /gm = 2.63 k

ro =



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B

r

+

v

E

gmv

C


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4.31

IC = ISeVBE/VT

1 +

VCE

VA

IC,Total = nIC

= nISeVBE/VT 1 +

VCE

VA

gm,Total =IC

VBE

= nIS

VTeVBE/VT

nIC

VT

= ngm

= n 0.4435 S

IB,Total =1

IC,Total

r,Total =IB,TotalVBE

1

IC,Total

VT

1

=

nIC

VT

1

=r

n

=225.5

n(assuming = 100)

ro,Total =IC,Total

VCE1

IC,Total

VA

1

=VA

nIC

=ro

n

=693.8

n


B

r,Total

+

v

E

gm,Totalv

C

ro,Total


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4.32 (a)

VBE = VCE (for Q1 to operate at the edge of saturation)

VT ln(IC/IS) = VCC ICRC

IC = 885.7 A

VB = VBE = 728.5 mV

(b) Let IC

, VB

, VBE

, and VCE

correspond to the values where the collector-base junction is forwardbiased by 200 mV.

VBE = V

CE + 200 mV

VT ln(I

C/IS) = VCC I

CRC + 200 mV

IC = 984.4 A

VB = 731.3 mV

Thus, VB can increase by V

B VB = 2.8 mV if we allow soft saturation.


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4.34

VBE = VCC IBRB

VT ln(IC/IS) = VCC ICRB/

IC = 1.67 mA

VBC = VCC IBRB (VCC ICRC)

< 200 mV

ICRC IBRB < 200 mV

RC

47395606 fundamental of microelectronics bahzad razavi chapter 4 solution manual (1)

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